Question

Integral sen⁴ x dx...............

Answer 1

$\int\limits {sen^4 x} \, dx = \frac{1}{4} ( \frac{3}{2}-sen2x + \frac{1}{8} sen 4x) + C$

Answer 2

Resposta:$sen^4x=(\frac{1-cos(2x)}{2})^2\\ \\ sen^4x= \frac{1}{4} - \frac{2cos(2x)}{4} +\frac{cos^2(2x)}{4}\\ \\ \int sen^4x\, dx = \int \frac{1}{4} \, dx - \frac{1}{2} \int cos(2x)\, dx +\frac{1}{4} \int cos^2(2x)\, dx \\ \\ cos^2(2x)=\frac{1+cos^2(4x)}{2} \\ \\ \int \frac{1}{4} \, dx-\frac{1}{2} \int cos(2x) \, dx+\frac{1}{4} \int \frac{1}{2} \, dx+ \frac{1}{8} \int cos(4x)\, dx =\frac{x}{4}+\frac{1}{4} sen(2x)+\frac{x}{8} +\frac{1}{32} sen(4x)\\ \\ =\frac{3x}{8} -\frac{1}{4}sen(2x) + \frac{1}{32}sen(4x)+C$